Answer
$F = 880~N$
Work Step by Step
In part (a), we found that the force on the ladder from the window is $280~N$
Since the system is in equilibrium, the frictional force on the ladder from the ground must also have a magnitude of $F_x = 280~N$
The vertical component of the force on the ladder from the ground is equal in magnitude to the weight of the worker and the ladder.
We can find $F_y$:
$F_y = (75~kg+10~kg)(9.8~m/s^2) = 833~N$
We can find the magnitude of the force on the ladder from the ground:
$F = \sqrt{F_x^2+F_y^2}$
$F = \sqrt{(280~N)^2+(833~N)^2}$
$F = 880~N$
